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Consider you have a slab waveguide having n1=3.5 and n2=1.45. The value of critical angel is 24.5 degree approx. Similarly effective refractive index will be in between 1.45 and 3.5. How can we find the eigenvalue for this optical waveguide and plot in Matlab or some other software.

Thanks and hope it is clear.

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closed as off-topic by John Rennie, Manishearth Jun 26 '13 at 0:23

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This paper could help you. For a numerical example, see formulae $1.83$ to $1.93$ pages $9,10$. –  Trimok Jun 25 '13 at 10:35

1 Answer 1

The slab waveguide is fully analysed ($i.e.$ full solutions of Maxwell's Equations are found) in Chapter 12 of Snyder and Love "Optical Waveguide Theory", Chapman and Hall 1983. A sketch of the solution is as follows:

One assumes pairs of plane wave solutions in the core and lone plane evanescent waves in the cladding: with the axial direction as $z$ ($i.e.$ the direction of no variation in refractive index, so that the waveguide is invariant under a translation along $z$) and $x$ in the transverse direction the solutions are of the following forms, where $\psi_j$ are the Cartesian components of the $\mathbf{E}$ and $\mathbf{H}$ vectors:

  1. In the core $\psi_j \propto \exp\left(i\, \beta \, z\right)\left[A_j \exp\left(i\,\sqrt{k^2 n_1^2 - \beta^2}\,x\right) + B_j \exp\left(-i\,\sqrt{k^2 n_1^2 - \beta^2}\,x\right)\right]$, where $\beta$ is the sought eigenvalue;

  2. In the cladding as $x\rightarrow\infty$ we have $\psi_j \propto \exp\left(i\, \beta \, z\right) \exp\left(-\sqrt{\beta^2 - k^2 n_2^2}\,x\right)$; and

  3. In the cladding as $x\rightarrow-\infty$ we have $\psi_j \propto \exp\left(i\, \beta \, z\right) \exp\left(\sqrt{\beta^2 - k^2 n_2^2}\,x\right)$

By imposing the wonted boundary conditions (continuity of tangential $\mathbf{E}$ and $\mathbf{H}$ components at the core-cladding interfaces), you'll find an eigenvalue equation for $\beta$. You'll end up with a homogeneous matrix equation of the form $\mathbf{M}(\beta)\, \mathbf{a} = \mathbf{0}$ where $\mathbf{a}$ is the column vector of the undetermined co-efficients $A_j$, $B_j$ in your assumed solutions for $\mathbf{E}$ and $\mathbf{H}$. Don't forget to impose the condition $\nabla \wedge \mathbf{E} = i \omega \mu \mathbf{H}$ ! (this will be part of the $\mathbf{M}$ matrix). Then the eigenvalue equation is a transcendental equation $\det \mathbf{M}(\beta) = 0$ in $\beta$.

Also, you get two classes of solutions: transverse electric (TE) fields, where there is an $x$ electric field component, no $z$ electric field component ($i.e.$ the electric field is wholly transverse, and the magnetic field has both $x$ and $z$ components. The second class is transverse magnetic (TM) fields, with the obvious meaning.

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